I know there are some pretty smart people browsing these forums, and I need a bit of quick help with physics. I’m working on an assignment for my AP Physics class through webassign, which is basically a way that teachers can put assignments online for students and they can fill out the homework on the internet and submit it for instant grading. Unfortunately they can also set the time it is due, and mine happens to be due in about 15 minutes.
I have been working on the last problem for about an hour and a half, and just can’t figure it out.
If you would be so kind to explain it to me, I would be very grateful.
Question:
A bicyclist is finishing his repair of a flat tire when a friend rides by at [color=#dd0000]3.8[/color] m/s. Two seconds later, the bicyclist hops on his bike and accelerates at [color=#dd0000]2.0[/color] m/s2 until he catches his friend.
Here’s the equation I have so far:
Nevermind about the urgent part, the deadline has passed now, but I would still appreciate if someone could explain this to me so I will know how to do problems like this in the future.
A bicyclist is finishing his repair of a flat tire when a friend rides by at 3.8 m/s. Two seconds later, the bicyclist hops on his bike and accelerates at 2.0 m/s2 until he catches his friend.
What is the question???
The “question” you have posted is a statement
Well, after time t, both bikes have travelled the same distance D.
After time T, bike 1 has travelled D = v01t+.5a1t²
After time T, bike 2 has travelled D = v02t+.5a2t² + 2v02 (he’s travelled 2 secondes at the speed of v02 hence the 2v02).
BUT when bike 1 starts, it’s speed it null, so v01 = 0, and since there’s no mention of bike 2 excelerating, we can assume a02 = 0 SO:
also, sorry about forgetting the question, you guys assumed right, they wanted to know how long it would take for the two guys to be at the same distance away.
